Thinking of arbitrary tensor products of rings, $A=\otimes_i A_i$ ($i\in I$, an arbitrary index set), I have recently realized that $Spec(A)$ should be the product of the schemes $Spec(A_i)$, a priori in the category of affine schemes, but actually in the category of schemes, thanks to the string of equalities (where $X$ is a not necessarily affine scheme)

Since this looks a little too easy, I was not quite convinced it was correct but a very reliable colleague of mine reassured me by explaining that the correct categorical interpretation of the more down to earth formula above is that the the category of affine schemes is a reflexive subcategory of the category of schemes.
(Naturally the incredibly category-savvy readers here know that perfectly well, but I didn't at all.)

And now I am stumped: I had always assumed that infinite products of schemes don't exist and I realize I have no idea why I thought so!

Since I am neither a psychologist nor a sociologist, arguments like "it would be mentioned in EGA if they always existed " don't particularly appeal to me and I would be very grateful if some reader could explain to me what is known about these infinite products.

representable by a scheme? If the $X_i$ are all affine, the answer is yes, as explained in the statement of the question. More generally, any filtered inverse system of schemes with essentially affine transition maps has an inverse limit in the category of schemes (this is in EGA IV.8). The topology in that case is the inverse limit topology, by the way.

It is easy to come up with examples of infinite products of non-separated schemes that are not representable by schemes. This is because any scheme has a locally closed diagonal. In other words, if $Y \rightrightarrows Z$ is a pair of maps of schemes then the locus in $Y$ where the two maps coincide is locally closed in $Y$.

Suppose $Z$ is the affine line with a doubled origin. Every distinguished open subset of an affine scheme $Y$ occurs as the locus where two maps $Y \rightrightarrows Z$ agree. Let $X = \prod_{i = 1}^\infty Z$. Every countable intersection of distinguished open subsets of $Y$ occurs as the locus where two maps $Y \rightarrow X$ agree. Not every countable intersection of open subsets is locally closed, however, so $X$ cannot be a scheme.

Since the diagonal of an infinite product of separated schemes is closed, a more interesting question is whether an infinite product of separated schemes can be representable by a scheme. Ilya's example demonstrates that the answer is no.

Let $Z = \mathbf{A}^2 - 0$. This represents the functor that sends $Spec A$ to the set of pairs $(x,y) \in A^2$ generating the unit ideal. The infinite product $X = \prod_{i = 1}^\infty Z$ represents the functor sending $A$ to the set of infinite collections of pairs $(x_i, y_i)$ generating the unit ideal. Let $B$ be the ring $\mathbf{Z}[x_i, y_i, a_i, b_i]_{i = 1}^\infty / (a_i x_i + b_i y_i = 1)$. There is an obvious map $Spec B \rightarrow X$. Any (nonempty) open subfunctor $U$ of $X$ determines an open subfunctor of $Spec B$, and this must contain a distinguished open subset defined by the invertibility of some $f \in B$. Since $f$ can involve at most finitely many of the variables, the open subset determined by $f$ must contain the pre-image of some open subset $U'$ in $\prod_{i \in I} Z$ for some finite set $I$. Let $I'$ be the complement of $I$. If we choose a closed point $t$ of $U'$ then $U$ contains the pre-image of $t$ as a closed subfunctor. Since the pre-image of $t$ is $\prod_{i \in I'} Z \cong X$ this shows that any open subfunctor of $X$ contains $X$ as a closed subfunctor.

In particular, if $X$ is a scheme, any non-empty open affine contains a scheme isomorphic to $X$ as a closed subscheme. A closed subscheme of an affine scheme is affine, so if $X$ is a scheme it is affine.

Now we just have to show $X$ is not an affine scheme. It is a subfunctor of $W = \prod_{i = 1}^\infty \mathbf{A}^2$, so if $X$ is an affine scheme, it is locally closed in $W$. Since $X$ is not contained in any closed subset of $W$ except $W$ itself, this means that $X$ is open in $W$. But then $X$ can be defined in $W$ using only finitely many of the variables, which is impossible.

Edit: Laurent Moret-Bailly pointed out in the comments below that my argument above for this last point doesn't make sense. Here is a revision: Suppose to the contrary that $X$ is an affine scheme. Then the morphism $p : X \rightarrow X$ that projects off a single factor is an affine morphism. If we restrict this map to a closed fiber then we recover the projection from $Z$ to a point, which is certainly not affine. Therefore $X$ could not have been affine in the first place.

Nice answer! I don't follow the last paragraph. You show any open subset $U$ of $X$ must contain a product of the form $\prod_{i \in I} Z$. At this point, you are basically done. But I don't understand the particular way you finish the proof. Why do you say "if $X$ is a scheme then it is an affine scheme."?
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David SpeyerDec 17 '09 at 12:10

I added some words to clarify this. The point is that $X$ has a nonempty open affine containing a closed subscheme isomorphic to $X$.
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Jonathan WiseDec 17 '09 at 16:22

Thank you for this masterful answer: you seem to be completely at home with EGA ! I never hoped to get such a prompt, clear and complete answer when I posted my question. I wish you all the success you obviously deserve.
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Georges ElencwajgDec 17 '09 at 22:50

Dear Jonathan, may I exploit your impressive expertise again? Are there classes of non affine schemes whose infinite products exist as schemes? Or does a modification of your proof show that infinite products of non-affine schemes never are schemes? Is, for example, a denumerable product of projective lines a scheme ? Very friendly, Georges.
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Georges ElencwajgDec 19 '09 at 10:30

Here's a guess at what goes wrong for general schemes. For simplicity, let X be a non-affine scheme; say it's the union of two affines: A1 and A2 (although I'm actually thinking of $\mathbb A^2 \backslash {0}$), and let's try to define the product $Y=\prod_{i=1}^\infty X$.

Well, we should be able to describe Y as a union of affines (that are glued along some maps). What should these be? There are two "obvious answers". If we carry over our intuition from topology, the natural building blocks should have the form $U_1 \times U_2 \times \ldots$ where each $U_i$ is one of $A_1, A_2$, or $X$ and all but finitely many $U_i$-s are equal to $X$. However, these products are not affine (they aren't really defined as schemes, but since $X$ is not affine, they are even "intuitively" not affine).

The second "obvious answer" would be to take products $U_1 \times U_2 \times \ldots$ where each $U_i$ is either $A_1$ or $A_2$. These would be affine, but this feels like a wrong answer: it would be like using the box topology on an infinite product. They shouldn't even be open in Y (I know, this is rather far-fetched since Y is not yet defined). Also, if you tried to glue Y out of these, I doubt you'd be able to define gluing maps (they are maps from an infinite product to an infinite product - I feel this is bad, but can categorically minded people confirm?).

So far, I don't have an actual proof that the second answer is bad, or that you couldn't define Y with some other affines, but I think there should be a good reason (the same reason as to why we use the product topology for topological spaces, though it eludes me at the moment).

Why would we want it to be a topological product at all? The topology on $\mathbb{A}^2$ is not the product topology $\mathbb{A}^1\times\mathbb{A}^1$. The thing you claim should be a basis for the topology of $Y$ just isn't, even in the finite product case.
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Charles SiegelDec 17 '09 at 2:05

@Charles: I feel that my answer was badly written, so I rewrote it. I think my point is much clearer now (although it's still far from a proof). Can you look at the new version and say whether you still disagree?
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Ilya GrigorievDec 17 '09 at 4:40

Here is a proof that the second answer is bad. Let's take $X=\mathbb{P}^1$, with $U_1$ and $U_2$ the standard affines. Consider infinitely many maps from $\mathbb{P}^1$ to $X$, with $f_i(t) = t-i$. Then the universal property of products gives us a map $F:\mathbb{P}^1 \to X \times X \times \cdots$. But $F^{-1}(U_2 \times U_2 \times \cdots)$ is the complement of infinitely many closed points, which is not an open set, contradicting that $F$ is continuous.
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David SpeyerDec 17 '09 at 11:59

Here is another example with a rigorous proof (which is a collaboration with "owk").

Example. Let $R$ be a discrete valuation ring, $I$ an infinite set. Glue two copies of $\text{Spec}(R)$ along the generic point to get a $R$-scheme $X$. Then in the category of $R$-schemes the power $X^I$ does not exist.

Proof: Write $\text{Spec}(R) = \{\eta,\mathfrak{m}\}$, where $\eta$ is the generic point and $\mathfrak{m}$ is the special point. Let $K$ be the quotient field and $k$ the residue field of $R$. Assume $P = X^I$ exists in the category of $R$-schemes.

For an $R$-scheme $T$, a $T$-valued point of $X$ corresponds to an open covering $T = T_1 \cup T_2$ such that $T_1 \cap T_2 = T_{\eta}$. If we apply this to $K$-schemes or $k$-schemes, we see $X \times_R K = \text{Spec}(K)$ and $X \times_R k = \text{Spec}(k) \coprod \text{Spec}(k) = \text{Spec} k[x]/(x^2-x)$. Now the reduction $X(R) \to X(k)$ is bijective: It maps $(\text{Spec}(R),\{\eta\}), (\{\eta\},\text{Spec}(R))$ to $(\text{Spec}(k),\emptyset), (\emptyset,\text{Spec}(k))$. From $P(T)=X(T)^I$ we deduce that also $P(R) \to P(k)$ is bijective.

Since fibers may be described by fiber products and fiber products commute with fiber products by general nonsense, we get as $K$-schemes

$P_{\eta} = (X \times_R K)^I = \text{Spec}(K)^I = \text{Spec}(K)$.

Let us denote the unique point in $P_{\eta}$ also by $\eta$. As $k$-schemes, we get

We see that $P_{\mathfrak{m}}$ is homeomorphic to $\{0,1\}^I$, in particular it is not discrete. Remark that $P_{\mathfrak{m}}$ is not open in $P$ since otherwise we would get the contradiction $P(R)=\emptyset$. Also remark that $P_{\mathfrak{m}}$ may be identified with $P(k)$, on which $\text{Aut}(P)$ acts transitively.

Next we want to show that $\eta$ is a generic point of $P$. If not, let $U$ be a nonempty open subset of $U$ with $\eta \notin U$. Then $U \subseteq P_{\mathfrak{m}}$ and it follows that $P_{\mathfrak{m}}$ is the union of the $\sigma(U)$, $\sigma \in \text{Aut}(P)$, and therefore open, contradiction.

Since $P_{\mathfrak{m}}$ is not discrete, there is some nonempty open subset $\text{Spec}(A) \subseteq P$ which contains two points $p_1,p_2 \in P_{\mathfrak{m}}$. They induce $p_1,p_2 \in P(k) \cong P(R)$. Since $R$ is local, $p_1,p_2$ are induced by $p_1,p_2 \in \text{Spec}(A)(R)$. But now $\text{Spec}(A)(R) \subseteq \text{Spec}(A)(K) = P(K)= \{\eta\}$, thus $p_1=p_2$, contradiction. -qed

If you want a tensor product satisfying the isomorphism described, you can just define it as the inductive limit of all finite tensor products. For example, if you tensor $k[x_i]$ like this you really obtain k[x1,x2,x3,...]. It seems that this is a quite reasonable construction.

looks ok to me. and in a sense, ega does have this result: in any category, arbitrary limits can be made from fiber products and filtered limits (and the terminal object i guess, but let's forget about that), and in the category of schemes fiber products always exist and filtered limits exist when the transition maps are affine.

Yes, they do exist. In general, any "algebraic" category such as Group, Ring, Vect has all colimits, and in particular all coproducts. (The precise statement is: every category monadic over Set has all small colimits, assuming that Set satisfies the axiom of choice.) The coproduct of a family (A_i) of rings can be described just like the ordinary (finite) tensor product, but with the restriction that each element must be representable as a finite sum of finite products of elements of A_i's.
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Tom LeinsterDec 17 '09 at 19:13

To add a tiny bit to Tom's description: in general, an arbitrary coproduct is the filtered colimit of the directed system of finite coproducts. So you take the filtered colimit in commutative rings of finite tensor products. This filtered colimit is a reflection or lifting of the ordinary filtered colimit in the category of sets; this is generally true for categories of algebras for a finitary monad on $Set$, or for an algebraic theory given by finitary operations.
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Todd Trimble♦May 20 '11 at 13:18

For Ring I mean commutative by unity.
COnsider schemas like ringed spaces (this category is bifibrated on Topological spaces category, and its fibre over X is $Ring-Shv(X)^{op}$, the schemas category is a full subcategory) then general product exist (as ringed spaces) and then the topological base space is the spaces product (this follow from the costruction of the limits in a fibrated category).
This product (as ringed space) is a schema iff is locally a affine schemas. Considering the base topology of product and the fact that the product of affine schemas is a (affine) schema, follow that:

if almost all (all but finite) schemas are affine then the product of schemas exist as schema.

For try to generalizee we need study if (or when) the infinite product of open sets is a open in the category of the Sobre topological spaces (local spaces).

More in general? I do this following idea, no too sure:

In Hakim (TOpos annelles and schemas relative)
she realize a schemas $SPEC(R)$ (i.e. a ringed space locally like the "Spec" of a ring) associated to a ringed-space $R$, generalizing the costruction os $Spec(R)$ from a ring $R$, of course this construction is universal in some sense.
QUestion: Define this construction a categorical reflection (or coreflection)? If Yes we can costruct the prodoct of schemas from the prodoct as ringed-spaces and then take the "SPEC" of this product